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I am reading Deligne: Hodge III, and am puzzled by a certain statement in section 10. If anyone could give a reference or a hint for how to prove this, I would be grateful. Maybe it is obvious and I just don't see why.

We consider an extension $G$ of an abelian variety by a torus. Then Deligne claims that the kernel of the exponential map $Lie(G) \to G$ can be identified with $H_1(G, \mathbb{Z} )$. Why is this true?

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I don't really know much about abelian varieties, but I can offer a comment that might help. If $G = \mathbb{C}^g/\mathbb{Z}^{2g}$ is a complex torus, then its Lie algebra can be identified with $\mathbb{C}^g$ in such a way that the exponential map may be thought of as the quotient map $\mathbb{C}^g \to G$; in particular, the kernel of exp is $\mathbb{Z}^{2g}$. On the other hand, $G$ is homeomorphic to a product of $2g$ copies of the unit circle, so that $H_1(G,\mathbb{Z}) = \mathbb{Z}^{2g}$. – Faisal Mar 30 '11 at 20:33
up vote 13 down vote accepted

The exponential map realizes Lie(G) as the universal covering space of G, and the kernel is group of covering transformations. Thus the kernel is $\pi_1(G)$, which, being commutative in this case, equals $H_1(G,Z)$. (I assume we are over the complex numbers.)

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That's a much better answer than mine. – Simon Rose Mar 30 '11 at 20:34
Why is the exponential surjective/a group homomorphism? – Daniel Litt Mar 30 '11 at 20:35
@Daniel Litt:In this case the image of the exponential map is an open subroup of G and G is connected therefore it is onto. – Mohan Ramachandran Mar 30 '11 at 21:00
... and it's a group homomorphism because the Lie algebra is abelian. – José Figueroa-O'Farrill Mar 30 '11 at 21:01
Ah, of course; extensions of abelian lie algebras are abelian. – Daniel Litt Mar 30 '11 at 22:49

This morally sounds like it should be related to the exponential sequence in sheaf cohomology, but I don't immediately see how... Although, absent the torus (which I take as meaning $(\mathbb{C}^\times)^n$), the exponential sequence there reads $$ H^1(A, \mathbb{Z}) \to H^1(A, \mathcal{O})\ \xrightarrow{\exp}\ H^1(A, \mathcal{O}^\times) $$ and (at least over $\mathbb{C}$) we have that $H^1(A,\mathcal{O}) \cong H^1(A, \mathbb{R}) \cong T_0^*A = \mathfrak{a}^\vee$. I know this isn't exactly what you're looking for, but it looks too similar to be completely off base.

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